Question

Va rog cat de repede

Answer 1

$\displaystyle x^2-3x\leq 0\\\\x^2-3x=0 \\ \\ a=1,~b=-3,~c=0\\ \\ \Delta=b^2-4ac=(-3)^2-4 \cdot 1 \cdot 0=9-0=9>0\\ \\ x_1=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-(-3)-\sqrt{9} }{2 \cdot 1} =\frac{3-3}{2} =\frac{0}{2} =0 \\ \\ x_2=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-(-3)+\sqrt{9} }{2 \cdot 1}=\frac{3+3}{2} =\frac{6}{2} =3$

$\displaystyle \left\begin{array}{ccc}x\\x^2-3x\\\end{array}\right|\left\begin{array}{ccc}-\infty~~~~~~~~~~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~~~~~~~3~~~~~~~~~~~~~~~~~~~~+\infty\\+++++++++0--------0+++++++++\\\end{array}\right\\ \\ \\ \mathbf{x\in[0;3]}$